algebra

Let's find what value(s) of  x  make  f(x) = 8

 

f(x)  =  8

 

x² - 6x + 4  =  8

 

x² - 6x - 4  =  0

 

By the Quadratic formula,

 

$x\ =\ \frac{6\pm\sqrt{(-6)^2-4(1)(-4)}}{2(1)}\ =\ \frac{6\pm\sqrt{52}}{2}\ =\ \frac{6\pm2\sqrt{13}}{2}\ =\ 3\pm\sqrt{13}$

 

And so....

 

$f(3+\sqrt{13})=8$       and       $f(3-\sqrt{13})=8$

 

First let's plug in  3 + √13  for  x  into the function  g( f(x) )  =  2x + 3

 

$g(\ 8\ )\ =\ g(\ f(3+\sqrt{13})\ )\ =\ 2(3+\sqrt{13})+3\ =\ 9+2\sqrt{13}$

 

Next let's plug in  3 - √13

 

$g(\ 8\ )\ =\ g(\ f(3-\sqrt{13})\ )\ =\ 2(3-\sqrt{13})+3\ =\ 9-2\sqrt{13}$

 

And so the sum of all the possible values of  g(8) is:

 

$(9+2\sqrt{13})\ +\ (9-2\sqrt{13})\ =\ 18$ .